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 Author Topic: Coulombs law and motion.  (Read 6651 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Mardoxx
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 « Embed this message on: April 22, 2010, 12:48:56 am » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

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Imagine two charged particles are placed a small distance, say 1 mm, apart and held there. The charge on each of the particles  is 0.02C.

One of the particles is released. Describe the motion of the free particle and calculate the total work done.

so;
$q_1 = q_2 = 0.02 \mathrm{C}$
$r = 1\times10^{-3} \mathrm{m}$
$F(r) = \frac{k \cdot q_1 q_2}{r^2}\mathrm{N} \text{ where } k = \frac{1}{4\pi\epsilon_0}$ $= 8.9875517873681764\times10^9 \mathrm{N \cdot m^2\cdot C^{-2}}$

We know that as $r$ increases $F$ decreases so,
$F(r)$ > $F(r+\delta r)$

We shall assume it is $q_2$ that is released and $q_1$ is held by an imaginary force.
When $q_2$ is released a force of $F(r) = \frac{k \cdot 0.02 \times 0.02}{(1\times10^{-3})^2}\mathrm{N}$ is pushing it in the positive direction.

After a tiny bit of time, $\delta t$, the particle will have moved $\delta r$ and so the force will have decreased meaning that the acceleration of the particle will have decreased...

This is where I get stuck.
how can we express the particles motion if it has no set mass...?

I was thinking change in momentum, but that still depends on mass...

This is an interesting one!
Please can you give me a pointer or two?

Is it actually possible to work it out without mass???
 « Last Edit: April 22, 2010, 01:20:29 am by Mardoxx » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #1 on: April 22, 2010, 02:47:49 pm » posted from:Taipei,T'ai-pei,Taiwan

The force $\vec{F}=\frac{kq_1 q_2}{r^2}\hat{r}$
So the work done from $r_0$ to $r$ is

$W=\int \vec{F}\cdot d\vec{s}=\int_{r_0}^{r} \frac{kq_1 q_2}{r'^2}dr'=-\frac{kq_1 q_2}{r}|_{r_0}^{r}=kq_1 q_2 (\frac{1}{r_0}-\frac{1}{r})$
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Mardoxx
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 « Embed this message Reply #2 on: April 22, 2010, 06:30:20 pm » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

Thanks
but how can I describe its motion without a mass?

Also, 0.02C ... that's pretty huge to be held at 1mm!!! The initial force comes out as... 355920000N!!!!
 « Last Edit: April 22, 2010, 08:06:56 pm by Mardoxx » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #3 on: April 22, 2010, 09:06:04 pm » posted from:Taipei,T\'ai-pei,Taiwan

Yes. the mass m is needed.

Initial velocity is zero, so $W=\tfrac{1}{2}mv^2=kq_1 q_2 (\frac{1}{r_0}-\frac{1}{r})$

so $v(r)= \sqrt{\frac{2kq_1 q_2}{m}(\frac{1}{r_0}-\frac{1}{r})}$
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Mardoxx
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 « Embed this message Reply #4 on: April 22, 2010, 09:22:53 pm » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

but again, I need this so I can do
position_new = position_old + velocity*dtime

this is confusing me lots
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Fu-Kwun Hwang
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 « Embed this message Reply #5 on: April 22, 2010, 09:26:41 pm » posted from:Taipei,T\'ai-pei,Taiwan

You need to ask the one who gave you the question.
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Mardoxx
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 « Embed this message Reply #6 on: April 22, 2010, 10:24:50 pm » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

I think i've got it

if I use F=ma(r) I can get
a(r) = (k*q1*q2)/(m*r**2)

then if i use euler/RK I can get velocity and then distance at time t

actually, no I don't know
because still, doing this... itsays it would get to the moon in a few seconds - and I'm not sure that's right....
 « Last Edit: April 22, 2010, 11:55:41 pm by Mardoxx » Logged
siamon
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 « Embed this message Reply #7 on: April 10, 2015, 02:22:42 pm » posted from:,,Satellite Provider

Your physics simulations are so impressive . I downloaded EJS and tried to play with it. I have already seen you flash tutorial.  However, could you give me some suggestions  - where and how  can I get some books or web tutorials  to learn EJS? I just feel that it would take me longer time to play with this software and understand it without reading a book or a help manual.
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lookang
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http://weelookang.blogspot.com

 « Embed this message Reply #8 on: April 10, 2015, 04:44:49 pm » posted from:,,Satellite Provider

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DieterLowe
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 « Embed this message Reply #9 on: April 17, 2015, 06:18:38 pm » posted from:Kannur,Kerala,India

The quantitative expression for the effect of these three variables on electric force is known as Coulomb's law. Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

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Louise88
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 « Embed this message Reply #10 on: October 08, 2015, 04:27:09 pm » posted from:Bangalore,Karnataka,India

Thank you it is a relevant information.
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You cannot always have happiness but you can always give happiness. ..."Mother Teresa(1910-1997, Roman Catholic Missionary, 1979 Nobel Peace Prize)"